Reductions Costas Busch - LSU

There is a deterministic Turing machine Computable function : There is a deterministic Turing machine which for any input string computes and writes it on the tape Costas Busch - LSU

Problem is reduced to problem If we can solve problem then we can solve problem Costas Busch - LSU

function (reduction) such that: Definition: Language is reduced to language There is a computable function (reduction) such that: Costas Busch - LSU

If: Language is reduced to and language is decidable Theorem 1: If: Language is reduced to and language is decidable Then: is decidable Proof: Basic idea: Build the decider for using the decider for Costas Busch - LSU

Decider for Decider Reduction for From reduction: END OF PROOF Input YES Input string YES accept accept (halt) Decider for Reduction NO NO reject reject (halt) From reduction: END OF PROOF Costas Busch - LSU

Example: is reduced to: Costas Busch - LSU

We only need to construct: Reduction Turing Machine for reduction DFA Costas Busch - LSU

Let be the language of DFA Reduction Turing Machine for reduction DFA construct DFA by combining and so that: Costas Busch - LSU

Costas Busch - LSU

Decider for Reduction Input string Decider YES YES NO NO Costas Busch - LSU

If: Language is reduced to and language is undecidable Theorem 2: If: Language is reduced to and language is undecidable Then: is undecidable Proof: Suppose is decidable Using the decider for build the decider for Contradiction! Costas Busch - LSU

If is decidable then we can build: Decider for YES Input string YES accept accept Decider for (halt) Reduction NO NO reject reject (halt) CONTRADICTION! END OF PROOF Costas Busch - LSU

To prove that language is undecidable we only need to reduce Observation: To prove that language is undecidable we only need to reduce a known undecidable language to Costas Busch - LSU

while processing input string ? State-entry problem Input: Turing Machine State String Question: Does enter state while processing input string ? Corresponding language: (while processing) Costas Busch - LSU

(state-entry problem is unsolvable) Theorem: is undecidable (state-entry problem is unsolvable) Proof: Reduce (halting problem) to (state-entry problem) Costas Busch - LSU

Decider for Decider Reduction Given the reduction, A contradiction! Halting Problem Decider Decider for state-entry problem decider YES YES Decider Reduction NO NO Given the reduction, if is decidable, then is decidable A contradiction! since is undecidable Costas Busch - LSU

We only need to build the reduction: So that: Costas Busch - LSU

For the reduction, construct from : special halt state halting states A transition for every unused tape symbol of Costas Busch - LSU

special halt state halting states halts halts on state Costas Busch - LSU

Therefore: halts on input halts on state on input Equivalently: END OF PROOF Costas Busch - LSU

Blank-tape halting problem Input: Turing Machine Question: Does halt when started with a blank tape? Corresponding language: Costas Busch - LSU

Theorem: is undecidable Proof: Reduce (halting problem) to (blank-tape halting problem is unsolvable) Proof: Reduce (halting problem) to (blank-tape problem) Costas Busch - LSU

Decider for Decider Reduction Given the reduction, A contradiction! Halting Problem Decider Decider for blank-tape problem decider YES YES Decider Reduction NO NO Given the reduction, If is decidable, then is decidable A contradiction! since is undecidable Costas Busch - LSU

We only need to build the reduction: So that: Costas Busch - LSU

Construct from : no yes Run with input If halts then halts too Accept and halt no Tape is blank? yes Run Write on tape with input If halts then halts too Costas Busch - LSU

halts when started on blank tape Accept and halt no Tape is blank? yes Run Write on tape with input halts on input halts when started on blank tape Costas Busch - LSU

halts when started on blank tape halts on input halts when started on blank tape Equivalently: END OF PROOF Costas Busch - LSU

If: Language is reduced to and language is undecidable Theorem 3: If: Language is reduced to and language is undecidable Then: is undecidable Proof: Suppose is decidable Then is decidable Using the decider for build the decider for Contradiction! Costas Busch - LSU

Suppose is decidable Decider for reject accept (halt) (halt) Costas Busch - LSU

Suppose is decidable Then is decidable Decider for Decider for NO YES reject accept Decider for (halt) YES NO accept reject (halt) Costas Busch - LSU

If is decidable then we can build: Decider for YES Input string YES accept accept Decider for (halt) Reduction NO NO reject reject (halt) CONTRADICTION! Costas Busch - LSU

Alternatively: Decider for Decider for Reduction CONTRADICTION! NO Input string YES reject accept Decider for (halt) Reduction YES NO accept reject (halt) CONTRADICTION! END OF PROOF Costas Busch - LSU

To prove that language is undecidable we only need to reduce Observation: To prove that language is undecidable we only need to reduce a known undecidable language to or (Theorem 2) (Theorem 3) Costas Busch - LSU

Undecidable Problems for Turing Recognizable languages Let be a Turing-acceptable language is empty? is regular? has size 2? All these are undecidable problems Costas Busch - LSU

Let be a Turing-acceptable language is empty? is regular? has size 2? Costas Busch - LSU

Empty language problem Input: Turing Machine Question: Is empty? Corresponding language: Costas Busch - LSU

(empty language problem) Theorem: is undecidable (empty-language problem is unsolvable) Proof: Reduce (membership problem) to (empty language problem) Costas Busch - LSU

Decider for Decider Reduction Given the reduction, A contradiction! membership problem decider Decider for empty problem decider YES YES Decider Reduction NO NO Given the reduction, if is decidable, then is decidable A contradiction! since is undecidable Costas Busch - LSU

We only need to build the reduction: So that: Costas Busch - LSU

Construct from : Tape of input string yes yes Turing Machine Accept Write on tape, and accepts ? Simulate on input Costas Busch - LSU

Louisiana The only possible accepted string yes yes Turing Machine Write on tape, and accepts ? Simulate on input Costas Busch - LSU 42 42

yes yes accepts does not accept Turing Machine Accept Write on tape, and accepts ? Simulate on input Costas Busch - LSU 43 43 43

Therefore: accepts Equivalently: END OF PROOF Costas Busch - LSU

Let be a Turing-acceptable language is empty? is regular? has size 2? Costas Busch - LSU

Regular language problem Input: Turing Machine Question: Is a regular language? Corresponding language: Costas Busch - LSU

(regular language problem) Theorem: is undecidable (regular language problem is unsolvable) Proof: Reduce (membership problem) to (regular language problem) Costas Busch - LSU

Decider for Decider Reduction Given the reduction, A contradiction! membership problem decider Decider for regular problem decider YES YES Decider Reduction NO NO Given the reduction, If is decidable, then is decidable A contradiction! since is undecidable Costas Busch - LSU

We only need to build the reduction: So that: Costas Busch - LSU

Construct from : Tape of input string yes yes Turing Machine Accept Write on tape, and accepts ? Simulate on input Costas Busch - LSU 50 50 50 50

yes yes not regular accepts does not accept regular Turing Machine Write on tape, and accepts ? Simulate on input Costas Busch - LSU 51 51 51 51 51

Therefore: accepts is not regular Equivalently: END OF PROOF Costas Busch - LSU

Let be a Turing-acceptable language is empty? is regular? has size 2? Costas Busch - LSU

Does have size 2 (two strings)? Size2 language problem Input: Turing Machine Question: Does have size 2 (two strings)? Corresponding language: Costas Busch - LSU

(size 2 language problem) Theorem: is undecidable (size2 language problem is unsolvable) Proof: Reduce (membership problem) to (size 2 language problem) Costas Busch - LSU

Decider for Decider Reduction Given the reduction, A contradiction! membership problem decider Decider for size2 problem decider YES YES Decider Reduction NO NO Given the reduction, If is decidable, then is decidable A contradiction! since is undecidable Costas Busch - LSU

We only need to build the reduction: So that: Costas Busch - LSU

Construct from : Tape of input string yes yes Turing Machine Accept Write on tape, and accepts ? Simulate on input Costas Busch - LSU 58 58 58 58 58

yes yes 2 strings accepts does not accept 0 strings Turing Machine Write on tape, and accepts ? Simulate on input Costas Busch - LSU 59 59 59 59 59 59

Therefore: accepts has size 2 Equivalently: END OF PROOF Costas Busch - LSU